Download Solutions to Homework Set #7 Phys2414 – Fall 2005

Solution Set #7
1
Solutions to Homework Set #7
Phys2414 – Fall 2005
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1. GRR1 6.P.002. A sled is dragged along a horizontal path at a constant speed
of 1.5 m/s by a rope that is inclined at an angle of 30.0◦ with respect to the
horizontal (the figure below). The total weight of the sled is 470 N. The tension
in the rope is 230 N. How much work is done by the rope on the sled in a time
interval of 5.0 s?
The formula for calculating the work done on an object is
W = |F~ |d cos θ
where |F~ | is the magnitude of the force applied, d is the distance traveled, and cos θ is the
cosine of the angle between the force vector and the displacement vector. In this case the
displacement is horizontal and equal to the speed times the time interval given, i.e.
d = vt
The work done is then given by
W = |F~ | (vt) cos θ
W = ( 230 N) · (1.5 m/s) · ( 5.0 s) · cos (30.0◦ )
W = 1.49 kJ
2. GRR1 6.P.005. A barge of mass 5.0 × 104 kg is pulled along the Erie Canal
by two mules, walking along towpaths parallel to the canal on either side of it.
The ropes harnessed to the mules make angles of 45◦ to the canal. Each mule is
pulling on its rope with a force of 1.0 kN. How much work is done on the barge
by both of these mules together as they pull the barge 130 m along the canal?
The total work done is just the sum of the work done by each individual force.
Wtotal =
X
i
Wi
Solution Set #7
2
The work done by a force is given by
WF = |F~ |d cos θ
So the work done by a single mule is given by
Wmule = (1.0 kN) · ( 130 m) · cos (45◦ ) = 92 kN
So the total work done is just the work done by each mule
Wtotal = Wmule + Wmule = 2Wmule
Wtotal = 184 kN
3. GRR1 6.P.012. A plane weighing 220 kN (25 tons) lands on an aircraft
carrier. The plane is moving horizontally at 64 m/s ( 143 mi/h) when its
tailhook grabs hold of the arresting cables. The cables bring the plane to a stop
in a distance of 83 m.
(a) How much work is done on the plane by the arresting cables?
Work total work done is equal to the change in kinetic energy. Since the arresting cables
are the only force that does work on the plane, then the work done by the arresting cables
is the total work. So the work done by the arresting cables is equal to the change in kinetic
energy.
W = ∆K
1 2 1 2
∆K = Kf − Ki = mvf − mvi
2
2
m = W/g
1
1
∆K = m(0)2 − ( 220 kN/9.81 m/s2 ) · ( 64 m/s)2
2
2
W = -45.9 MJ
(b) What is the force (assumed constant) exerted on the plane by the cables?
The work done by a force is given by:
W = |F~ |d cos θ
Solving for the force |F~ | we have the following formula. We plug in our previously calculated
value for work, and because the cables pull on the plane in the opposite direction of the
Solution Set #7
3
plane’s motion we use 180◦ for the angle.
|F~ | =
-45.9 MJ
W
=
d cos θ
( 83 m cos (180◦ )
|F~ | = 553 kN
4. GRR1 6.P.021. Justin moves a desk 4.8 m across a level floor by pushing
on it with a constant horizontal force of 310 N. (It slides for a negligibly small
distance before coming to a stop when the force is removed.) Then, changing his
mind, he moves it back to its starting point, again by pushing with a constant
force of 310 N.
(a) What is the change in the desk’s gravitational potential energy during the round-trip?
The change in gravitational potential energy is given by the following formula. Since the
initial and final positions are the same (hf −hi ) = 0. So the change in gravitational potential
energy is.
∆Ug = mghf − mghi = mg(hf − hi ) = mg(0) = 0
(b) How much work has Justin done on the desk?
The total work is the sum of all the individual works
Wtotal =
X
Wi
i
The work done on a single trip is then:
Wsingletrip = |F~ |d cos θ = ( 310 N) · ( 4.8 m) cos (0◦ )
Wsingletrip = 1.49 kJ
So the total work done is just the sum of the two single trips.
Wtotal = 2WSingletrip = 2.98 kJ
(c) If the work done by Justin is not equal to the change in gravitational potential energy
of the desk, then where has the energy gone?
Friction is a non-conservative force, so the mechanical energy is not conserved. The energy
is dissipated in the form of heat
Solution Set #7
4
5. GRR1 6.P.028. When a 0.20 kg mass is suspended from a vertically hanging
spring, it stretches the spring from its original length of 3.0 cm to a total
length of 7.0 cm. The spring with the same mass attached is then placed on
a horizontal frictionless surface. The mass is pulled so that the spring stretches
to a total length of 10.0 cm; then the mass is released and it oscillates back and
forth (the figure below). What is the maximum speed of the mass as it oscillates?
This problem has two parts, one where the spring is hanging from the ceiling and another
where the spring is on a frictionless horizontal surface. We use the information of how the
spring behaves as you hang is from a ceiling in to get the spring constant k. We use Newtons
second law on the mass to do this.
X
F~ = m~a
The only forces acting on the spring are the force of the spring, and weight. Solving the
equation for k gives us:
Fspring − mg = 0
(.2 kg) · (9.81 m/s2 )
mg
=
k=
x
( 7.0 − 3.0 ) cm
k = 49 N/m
In the second part we use the conservation of energy to find the maximum velocity of the
block. The conservation of energy states:
∆K + ∆Us = WN C
Where ∆K is the change in kinetic energy, ∆U is the change in potential energy, and WN C
is the work done by non-conservative forces. We then plug in K = 1/2mv 2 and Us = 1/2kx2 ,
and since there is no friction there is no work done by non-conservative forces or W N C is
zero.
1 2 1 2 1 2 1
mv − mv + kx − mx2i = 0
2 f 2 i 2 f 2
We know the initial velocity is zero, and solving for the final velocity we get.
k 2
(xi − x2f )
m
s
k 2
vf =
(x − x2f )
m i
vf2 =
Solution Set #7
5
So if we want the final velocity to be a maximum, then obviously xf must be zero. We can
then plug in our value for the spring constant k we calculated, the mass, and the initial
stretch of the spring (which would be the stretched length 10 cm minus the unstretched
length 3 cm.
vf =
vf =
v
u
u 49 N/m
t
0.2 kg
s
k
(xi )
m
((10 − 3.0 ) cm)
vf = 1.1 m/s
6. GRR1 6.P.029. A cart starts from position 4 in the figure below with a
velocity of 13 m/s to the left. Find the speed with which the cart reaches
positions 1, 2, and 3. Neglect friction.
We want to start with the conservation of energy.
∆K + ∆Ug = WN C
We then plug in K = 1/2mv 2 and U = mgh, and use the fact that there is no friction so
WN C = 0. We can then solve for vf .
1 2 1 2
mv − mv + mghf − mghi = 0
2 f 2 i q
vf = vi2 + 2g(hi − hf )
speed at position 1
For these parts we just plug in the values of initial velocity vi , initial height hi and final
height hf .
vf =
q
( 13 m/s)2 + 2(9.81 m/s2 ) · (20 − 0) m
vf = 23.7 m/s
speed at position 2
vf =
q
( 13 m/s)2 + 2(9.81 m/s2 ) · (20 − 15) m
vf = 16.3 m/s
Solution Set #7
6
speed at position 3
vf =
q
( 13 m/s)2 + 2(9.81 m/s2 ) · (20 − 10) m
vf = 19.1 m/s
7. GRR1 6.P.034. An object slides down an inclined plane of angle 30.0 ◦ and
of incline length 2.0 m. If the initial speed of the object is 6.0 m/s directed
down the incline, what is the speed at the bottom? Neglect friction.
We can do this problem like problem # 6. We start with the conservation of energy, plug in
expressions for kinetic energy K and potential energy U . There is no friction so W N C = 0.
We then solve for vf .
∆K + ∆Ug = WN C
1 2 1 2
mv − mv + mghf − mghi = 0
2 f 2 i q
vf = vi2 + 2g(hi − hf )
Next we have to realize that the change in height (hi − hf ) is related to the distance the
block travels down the incline d.
³
´
hi − hf = d sin θ = ( 2.0 m) · sin 30 ◦ = 1 m
Finally we just plug in values.
vf =
q
(6.0 m/s)2 + 2(9.81 m/s2 ) · ( 1 m)
vf = 7.46 m/s
8. GRR1 6.P.039. A spring with k = 10.0 N/m is at the base of a frictionless
30.0 inclined plane. A 0.50 kg object is pressed against the spring, compressing
it 0.2 m from its equilibrium position. The object is then released. If the object
is not attached to the spring, how far up the incline does it travel before coming
to rest and then sliding back down? (See the figure below.)
We start with the conservation of energy equation, where for this problem we include both
a gravitational potential energy term Ug and a spring potential energy term Us .
∆K + ∆Ug + ∆Us = WN C
Solution Set #7
7
Without friction we know that WN C = 0. We also know that since our initial and final
velocities are zero, then the total change in kinetic energy is zero.
1
1
∆K = mvf2 − mvi2 = 0 − 0 = 0
2
2
Plugging in expressions for ∆Ug and ∆Us we get the following formula.
1
1
mghf − mghi + kx2f − kx2i = 0
2
2
We then have to relate the change in height to the distance traveled up the incline d:
hf − hi = d sin θ
We then just solve for the distance up the incline d.
d=
kx2i
( 10 N/m)( 0.2 m)2
=
2mg sin θ
2 · (0.50 kg) · (9.81 m/s2 ) · sin (30◦ )
d = 0.082 m
9. GRR1 6.P.044. A spring gun (k = 28 N/m) is used to shoot a 56 g ball
horizontally. Initially the spring is compressed by 19 cm. The ball loses contact
with the spring and leaves the gun when the spring is still compressed by 12 cm.
What is the speed of the ball when it hits the ground, 1.4 m below the spring
gun?
We start with the conservation of energy. Then plug in the expressions for ∆K, ∆U g , and
∆Us , and since there is no friction or air resistance we can set WN C = 0.
∆K + ∆Ug + ∆Us = WN C
1
1
1 2 1 2
mvf − mvi + mghf − mghi + kx2f − kx2i = 0
2
2
2
2
Since the ball starts from rest we set the initial velocity vi to zero, and solve for the final
velocity. We can then plug in the given values for the change in height hi − hf , the spring
constant k, the mass m, and the initial and final values of the spring compressions xi and
xf .
vf =
vf =
v
u
u
t
2 · (9.81 [m/s2 )( 1.4 − 0) m +
s
2g(hi − hf ) +
k 2
(x − x2f )
m i
28 N/m
(( .19 m)2 − ( .12 m)2 )
.056 kg
vf = 6.19 m/s
Solution Set #7
8
10. GRR1 6.P.057. The power output of a cyclist moving at a constant speed of
5.0 m/s on a level road is 110 W.
(a) What is the force exerted on the cyclist and the bicycle by the air?
Power is given by the formula P = F v, where P is power, F is the force, and v is the velocity.
So using that formula and solving for F gives:
P = Fv
F = P/v = 110 W/ 5.0 m/s
F = 22 N
(b) By bending low over the handlebars, the cyclist reduces the air resistance to 16 N.
If she maintains a power output of 110 W, what will her speed be?
Solving the power formula for velocity gives:
P = Fv
v = P/F = 110 W/ 26 N
F = 6.9 m/s
11. GRR1 6.P.078. [299744] A 0.50 kg block, starting at rest, slides down a 30.0
incline with kinetic friction coefficient 0.28 (the figure below). After sliding
83 cm down the incline, it slides across a frictionless horizontal surface and
encounters a spring (k = 33 N/m).
(a) What is the maximum compression of the spring?
We should use the conservation of energy to solve this problem, so we start out with the
conservation of energy equation.
∆K + ∆Ug + ∆Us = WN C
First we look at the change in kinetic energy ∆K. Since the initial and final velocities are
both zero, then the change in kinetic energy is also zero.
1
1
∆K = mvf2 − mvi2 = 0 − 0 = 0
2
2
We then plug in expressions for ∆Ug and ∆Us , and since there is friction we replace WN C
with the work done by friction Wf .
1
1
mghf − mghi + kx2f − kx2i = Wf
2
2
Solution Set #7
9
But we need to figure out what the work done by friction is. The formula for work done by
a force is:
Wf = |F~f |d cos (180◦ )
But now we need to find the force of friction. We start by use the expression the expression
for force of kinetic friction (kinetic since the block is moving). Ff = µk FN . Drawing a
free body diagram, and setting the axis up along the plane we get for the sum of forces
perpendicular to the plane.
FN − mg cos (30◦ ) = 0
So the force of friction is then given by the following formula, and the work done by friction
is then similarly given by the next formula.
Ff = µk mg cos (30◦ )
Wf = ( 0.28 ) · (0.50 kg) · (9.81 m/s2 ) cos (30◦ ) · ( 83 cm) · cos (180◦ ) = -0.99 J
Plugging this into our equation for conservation of energy, and setting xi = 0 and (hi −hf ) =
d cos(30◦ ), we get:
1 2
kx = mgd sin (30◦ ) + Wf = (0.50 kg) · (9.81 m/s2 ) · ( 83 cm) sin (30◦ ) − -0.99 J
2 f
1 2
kx = 1.04 J
2 f
q
xf =
2 · ( 1.04 J)/( 33 N/m)
xf = 25 cm
(b) After the compression of part (a), the spring rebounds and shoots the block back up
the incline. How far along the incline does the block travel before coming to rest?
Again we want to use the conservation of energy to solve this problem. And again for the
same reason as in part a we want to set ∆K = 0.
∆K + ∆Ug + ∆Us = WN C
1
1
∆K = mvf2 − mvi2 = 0 − 0 = 0
2
2
So plugging in the usual expressions for ∆Ug and ∆Us and setting WN C = Wf we get:
1
1
mghf − mghi + kx2f − kx2i = Wf
2
2
Solution Set #7
10
Again we replace the change in height with the distance up the plane times the sine of the
angle of the plane, and we use the formula for work done by a force.
hf − hi = d sin θ
Wf = |F~f |d cos (180◦ )
And finally we solve for d the distance up the plane the block slides.
1
mgd sin θ + |F~f |d = kx2i
2
1
2
kx
1.04
J
i
2
d=
=
mg(sin θ + µk cos θ)
(0.50 kg) · (9.81 m/s2 ) · (sin (30◦ ) + 0.28 cos (30◦ ))
d = 28.5 cm
12. GRR1 6.TB.042. [219697] A roller coaster car (mass = 988 kg including
passengers) is about to roll down a track (the figure below ). The diameter of
the circular loop is 20.0 m and the car starts out from rest 40.0 m above the
lowest point of the track. Ignore friction and air resistance and assume g = 9.81
m/s2.
(a) At what speed does the car reach the top of the loop?
We start with the conservation of energy equation, and plug in the usual expressions for ∆K
and ∆Ug , and set WN C = 0 since there is no friction.
∆K + ∆Ug = WN C
1 2 1 2
mv − mv + mghf − mghi = 0
2 f 2 i
We then solve for the final velocity.
vf =
vf =
q
q
vi2 + 2g(hi − hf )
(0)2 + 2(9.81 m/s)(40 − 20) m
vf = 19.8 m/s
(b) What is the force exerted on the car by the track at the top of the loop?
The question ask about a force, so we start with a free body diagram, the free body diagram
only has two forces: mg pointing straight down, and FN also pointing down since the track
is above the cart at the top of the loop. Newton’s Second law then states:
X
F~ = m~a
Solution Set #7
11
But we know that the acceleration is the centripetal acceleration, so a = v 2 /r. Solving for
FN gives:
mv 2
FN + mg =
r
FN =
(988 kg) · (19.8 m/s)2
mv 2
− mg =
− (988 kg) · (9.81 m/s)
r
10 m
FN = 29.1 kN
(c) From what minimum height above the bottom of the loop can the car be released so
that it does not lose contact with the track at the top of the loop?
To do this problem we first want to find the minimum velocity the car must have to stay on
the track on the top of the loop, and then use the conservation of energy to find from what
height the cart must be released so that it reaches that velocity.
To find from what the minimum velocity needed for the cart to stay on the track we must
use the newtons second law again, but let the normal force just go to zero.
FN + mg =
mv 2
r
So letting FN = 0 and solving for v 2 gives us:
v 2 = gr
Now we want to use the conservation of energy to find hi , if vf2 = gr.
∆K + ∆Ug = WN C
1 2 1 2
mv − mv + mghf − mghi = 0
2 f 2 i
1
1
mghi = mghf + mvf2 = mgh + mgr
2
2
1
1
hi = hf + r = 20 m + 10 m
2
2
hi = 25 m